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Q11:
Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?
Solution :
Given: An Arithmetic Progression (AP) defined by the sequence: $3, 15, 27, 39, \dots$
To Find: The term number $n$ such that the $n^{th}$ term ($a_n$) is $132$ more than the $54^{th}$ term ($a_{54}$).
Step 1: Identify the parameters of the Arithmetic Progression.
The general form of an AP is $a, a+d, a+2d, \dots$, where $a$ is the first term and $d$ is the common difference.
From the given sequence $3, 15, 27, 39, \dots$:
First term ($a$) = $3$
Common difference ($d$) = $a_2 - a_1 = 15 - 3 = 12$
[Verification: $27 - 15 = 12$ and $39 - 27 = 12$. Since the difference is constant, $d = 12$.]
Step 2: State the formula for the $n^{th}$ term of an AP.
The formula for the $n^{th}$ term of an AP is given by:
$a_n = a + (n - 1)d$
Step 3: Calculate the $54^{th}$ term ($a_{54}$).
Using the formula $a_n = a + (n - 1)d$ with $n = 54$, $a = 3$, and $d = 12$:
$a_{54} = 3 + (54 - 1) \times 12$
$a_{54} = 3 + (53 \times 12)$
$a_{54} = 3 + 636$
$a_{54} = 639$
Step 4: Formulate the equation based on the problem statement.
The problem states that the $n^{th}$ term is $132$ more than the $54^{th}$ term:
$a_n = a_{54} + 132$
Substituting the value of $a_{54}$ found in Step 3:
$a_n = 639 + 132$
$a_n = 771$
Step 5: Solve for $n$ using the $n^{th}$ term formula.
Substitute $a_n = 771$, $a = 3$, and $d = 12$ into the general formula $a_n = a + (n - 1)d$:
$771 = 3 + (n - 1) \times 12$
Subtract $3$ from both sides:
$771 - 3 = (n - 1) \times 12$
$768 = (n - 1) \times 12$
Divide both sides by $12$:
$\frac{768}{12} = n - 1$
$64 = n - 1$
Add $1$ to both sides:
$n = 64 + 1$
$n = 65$
Final Answer: The 65th term of the AP will be 132 more than its 54th term.
More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.2
- Q1(i): Fill in the blanks in the following table, given that $a$ is the first term, $d$ the common difference and $a_n$ the $n$th term of the AP: (i) $a=7, d=3, n=8, a_n=...$
- Q1(ii): Fill in the blanks in the following table, given that $a$ is the first term, $d$ the common difference and $a_n$ the $n$th term of the AP: (ii) $a=–18, d=..., n=10, a_n=0$
- Q1(iii): Fill in the blanks in the following table, given that $a$ is the first term, $d$ the common difference and $a_n$ the $n$th term of the AP: (iii) $a=..., d=–3, n=18, a_n=–5$
- Q1(iv): Fill in the blanks in the following table, given that $a$ is the first term, $d$ the common difference and $a_n$ the $n$th term of the AP: (iv) $a=–18.9, d=2.5, n=..., a_n=3.6$
- Q1(v): Fill in the blanks in the following table, given that $a$ is the first term, $d$ the common difference and $a_n$ the $n$th term of the AP: (v) $a=3.5, d=0, n=105, a_n=...$
- Q10: The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
- Q12: Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
- Q13: How many three-digit numbers are divisible by 7?
- Q14: How many multiples of 4 lie between 10 and 250?
- Q15: For what value of $n$, are the $n$th terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?
- Q16: Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
- Q17: Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.
- Q18: The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
- Q19: Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?
- Q2(i): Choose the correct choice in the following and justify : (i) 30th term of the AP: 10, 7, 4, . . . , is
- Q2(ii): Choose the correct choice in the following and justify : (ii) 11th term of the AP: – 3, $-\frac{1}{2}$, 2, . . ., is
- Q20: Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the $n$th week, her weekly savings become ₹ 20.75, find $n$.
- Q3(i): In the following APs, find the missing terms in the boxes : (i) 2, ☐, 26
- Q3(ii): In the following APs, find the missing terms in the boxes : (ii) ☐, 13, ☐, 3
- Q3(iii): In the following APs, find the missing terms in the boxes : (iii) 5, ☐, ☐, $9\frac{1}{2}$
- Q3(iv): In the following APs, find the missing terms in the boxes : (iv) – 4, ☐, ☐, ☐, ☐, 6
- Q3(v): In the following APs, find the missing terms in the boxes : (v) ☐, 38, ☐, ☐, ☐, – 22
- Q4: Which term of the AP : 3, 8, 13, 18, . . . ,is 78?
- Q5(i): Find the number of terms in each of the following APs : (i) 7, 13, 19, . . . , 205
- Q5(ii): Find the number of terms in each of the following APs : (ii) 18, $15\frac{1}{2}$, 13, . . . , – 47
- Q6: Check whether – 150 is a term of the AP : 11, 8, 5, 2 . . .
- Q7: Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
- Q8: An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
- Q9: If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?
CBSE Solutions for Class 10 Mathematics Arithmetic Progression
Chapters in CBSE - Class 10 Mathematics
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